If two rings have isomorphic lattices of ideals and one them is a domain, is the other necessarily a domain?
The rings in question are commutative and have an element $1$.
EDIT. I apologize for a lot of useless edits. I convinced myself that it was natural to require that our lattice isomorphism preserve arbitrary joins and meets, and that I had missed this subtlety; and I edited the question accordingly. Then Eric Wofsey observed that lattice isomorphisms automatically preserve arbitrary joins and meets. So I made at least two mistakes: (1) I failed to think of the completeness condition, (2) I thought that it was necessary to require it!
No. For instance, let $A$ be a valuation ring with value group $\mathbb{Q}$, and let $B=A/I$ where $I$ is the ideal of elements of valuation at least $1$. Ideals in $A$ canonically correspond to upward-closed subsets of $[0,\infty)\cap\mathbb{Q}$ with the inclusion order. Similarly, ideals in $B$ correspond to upward-closed subsets of $[0,1)\cap\mathbb{Q}$. Since $[0,\infty)\cap\mathbb{Q}$ and $[0,1)\cap\mathbb{Q}$ are order-isomorphic, so are the lattices of ideals in $A$ and $B$ (explicitly, they are each order-isomorphic to the Cantor set with a greatest element added on). However, $A$ is a domain and $B$ is not.